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2h
comment What is the intuition of why convergence in distribution does not imply convergence in probability
For convergence in probability, the sequence $X_i$ is a sequence of approximations to $X$, so there must be strong dependence (positive correlation) between them. For convergence in distribution, it's the distribution of $X_i$ that converges, like in the central limit theorem. It doesn't matter whether there is any dependence between the sequence $X_i$ and the limiting variable $X$.
12h
comment Different ways of giving away 35 coins to 5 people?
If each variable had to be the same then you get the unique solution $a=b=c=d=e=7$. More generally, if you're trying to solve a different problem, then the solution is different. As a rule of thumb.
13h
answered Different ways of giving away 35 coins to 5 people?
13h
answered What is the intuition of why convergence in distribution does not imply convergence in probability
15h
comment Order of Growth of a Sum
It's perfectly possible that the argument can be simplified. If you manage to simplify it, you are welcome to post your own answer. Note, however, that the asymptotics of your suggested bound is roughly $rk/\sqrt{2\pi(k/n)(1-k/n)} = rn\sqrt{k/2\pi(n-k)}$ which grows linearly with $r$.
15h
comment for $n > 1$ : odd , prove that $\Phi_{2n}(x) = \Phi_{n}(-x)$
This is just a hint, the rest you should do on your own.
1d
comment Counting the functions with f(i) ≤ f(i+1) for all i=1,..,n-1
I had better, if you want $f(i) < f(i+1)$ then only the identity function $f(i) = i$ works.
1d
answered for $n > 1$ : odd , prove that $\Phi_{2n}(x) = \Phi_{n}(-x)$
1d
answered “Notes on Set Theory” - Badly Written?
1d
answered A conjecture on binomial factors
1d
answered Counting the functions with f(i) ≤ f(i+1) for all i=1,..,n-1
2d
answered Order of Growth of a Sum
May
19
answered A convolution involving binomials
May
19
comment Factoring product of two primes from solutions of congruence
Right, that's the trick.
May
19
answered Factoring product of two primes from solutions of congruence
May
19
comment A recurrence for a combinatorial problem
No, my hint should be enough. Spend a few hours on this, and eventually you'll get it.
May
19
comment A recurrence for a combinatorial problem
This case doesn't happen.
May
19
comment how to simplify $\sqrt{\cos (x)} \sinh \left(\ln (2) x^{\cos(x)}\right)+\sqrt{\cos (x)} \cosh \left(\ln (2) x^{\cos(x)}\right)$
There's nothing there to solve. It's like asking how to solve $x^3$.
May
19
revised how to simplify $\sqrt{\cos (x)} \sinh \left(\ln (2) x^{\cos(x)}\right)+\sqrt{\cos (x)} \cosh \left(\ln (2) x^{\cos(x)}\right)$
added 5 characters in body; edited title
May
19
answered A recurrence for a combinatorial problem